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3 years ago

I need help pick only 4 questions

Mathematics

1 answer:

Tatiana [17]3 years ago

4 0

18. 0.0011
20. 1/625
19. 0.0011
21. 1/27

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Five years ago Sara's aunt was four times as old as Sara. If Sara's aunt is 42 years older than Sara, how old is Sara now? Pleas

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2 years ago

Explain how finding 7×20 is similar to finding 7×2,000 . then Find each product

3241004551 [841]

7x20=140. 7x2000=14000. The only thing that changes the product of these numbers is the amount of zeros behind the 2. Since the only two numbers that affect the answer is the 7 and the 2. (7x2=14). The number of zeros behind the 2 affect how many zeros will be included in the product.

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3 years ago

Read 2 more answers

Find the inverse of function f. f(x) = 1/2x +7 / <img src="https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B1%7D%7B2%7Dx%20%2B%207"

kvasek [131]

The answer is C.) f^-1 (x) = 2x- 14 / $\:f^{-1\:}\left(x\right)=2x\:-\:14$

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3 years ago

Simplyfy the given equation 5x+2(x-3)=-2(x-1)

Blizzard [7]

5x+2x−6=−2x+2

7x−6=−2x+2

7x=−2x+2+6

7x=−2x+8

7x+2x=8

9x=8

x=8/9

Hope this helps, have a blessed day! :-)

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3 years ago

Rewrite the system of linear equations as a matrix equation AX = B.

iren2701 [21]

Answer:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]$

And the vector of vriables (X) is:

$\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]$

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

$\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

In conclusion, AX=B is:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

7 0

3 years ago